114 V NORMED SPACES

10. SEPARABLE NORMED SPACES

(5.10.1) If in a normed space E there exists a total sequence (Section 5.4), E is
separable. Conversely., in a separable normed space E, there exists a total
sequence consisting of linearly independent vectors.

Suppose (an) is a total sequence, and let D be the set of all (finite) linear
combinations r^ + • • • -f rnan with rational coefficients (when E is a complex
vector space, by a "rational" scalar we mean a complex number a + //?,
with both a and /? rational). D is a denumerable set by (1.9.3) and (1.9.4).
As by definition the set L of all linear combinations of the an is dense in E,
all we have to prove is that D is dense in L, and as

||(i1fl1 + ••• .H^aJ-Oyii + ••• + rnan}\\ < fjA,- r,| • ||ay||,

this follows from (2.2.16).

Suppose conversely E is separable; we can of course suppose E is infinite
dimensional (otherwise any basis of E is already & finite total subset). Let
(an) be an infinite dense sequence of vectors of E. We define by induction a
subsequence (akn) having the property that it consists of linearly independent
vectors and that for any m^kn., am is a linear combination of akl, ..., akn.
To do this, we merely take for ki the first index for which an ^ 0, and for
kn+l the smallest index m > kn such that am is not in the subspace Vn gener-
ated by akl, ...,akn; such an index exists, otherwise, as VM is closed by
(5.9.2), Vn would contain the closure E of the set of all the an> contrary to
assumption. It is then clear that (aki) has the required properties, and is
obviously by construction a total sequence.

PROBLEM

Show that the spaces (c0) and I1 of Banach (Section 5.3, Problem 5, and Section 5.7,
Problem 1) are separable, but that the space 7°° (Section 5.7, Problem 1) is not separable.
(Show that in /°° there exists a nondenumerable family (XA) of points such that \\x* — x^ \\ = 1
for A ^ ^, using Problem 2(b) of Section 4.2, and (2.2.17).)here which is compact is finite dimen-